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AQA Mathematics > mechanics > Flashcards

mechanics Flashcards

1
Q

Hookes law equation

A

T = λx / l

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2
Q

assumptions for hookes law

A
  • the string/ spring is light
  • The system has no friction
  • The string / spring obeys hookes law
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3
Q

hookes law on two springs:

A
  1. draw a diagram for lengths, draw a diagram for tensions
  2. find the forces + equate if theyre in equilibrium
  3. apply hookes law to the tensions in the strings
  4. sub these tensions into the equations made in step 2
  5. solve
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4
Q

work done by a variable force in a string/ spring

A

λx²/2l

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5
Q

in perfectly elastic strings:

A

energy is conserved:

Final KE, GPE, EPE = Initial KE, GPE, EPE

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6
Q

if a non conservative force does work on a string

A

work done = change in energy

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7
Q

Solving vertical motion involving elastic forces using energy

A
  1. calculate loss/ gain in GPE and KE

2. use conservation of energy to work out λ or x

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8
Q

Solving vertical motion involving elastic forces using calculus

A
  1. Use F = ma where a is a derivative
  2. solve differential equation
  3. input initial variables
  4. input question conditions
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9
Q

all quantities can be expressed in terms of:

A

Mass
Length
Time

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10
Q

tangential speed, v =

A

rω

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11
Q

radial acceleration is

A

acceleration towards the centre of the circle

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12
Q

radial acceleration, a =

A

rω² = v²/r

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13
Q

Force towards the centre of the circle, F =

A

mrω² = mv²/r

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14
Q

Examples of centripetal force

A
  • Tension in a string
  • friction
  • A reaction force
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15
Q

In horizontal circles consider the forces:

A
  • forces towards/ away from the centre of the circle

- forces perpendicular to the plane of motion

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16
Q

working out speed in vertical circles

A

conservation of energy

GPE = KE

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17
Q

tangential acceleration =

A

ra

a is the angular acceleration

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18
Q

Constant angular acceleration with variable speed:

A 
Suvat equations where 
ω = v
ωo = u
angular acceleration, a = a
t = t
θ = s
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19
Q

If a particle just leaves its circular path then

A

R>0 T=0

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20
Q

solving a circular motion problem

A
  1. use forces (F=ma)
  2. energy analysis
  3. sub equations together and solve
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21
Q

impulse =

A

Change in momentum = force * time

Ft = m(v-u)

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22
Q

∫Fdt =

A

impulse

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23
Q

∫Fds =

A

work done

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24
Q

work done vector form

A

W = F.d

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25
Q

power vector form

A

P = F.v

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26
Q

Kinetic energy vector form

A

KE = ½mv.v

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27
Q

alternate form for acceleration

A

dv/dt
d²x/dt²
vdv/dx

28
Q

a force may be dependent on:

A
  • time F= f(t)
  • displacement F = f(x)
  • velocity F = f(v)
29
Q

including the effects of air resistance

A
  1. F = ma Where F is the sum of forces and a is a derivative
  2. separate variables
  3. integrate and add arbitrary constant
  4. use initial conditions to find constant
  5. input conditions given in question
30
Q

When no forces acting except friction then

A

KE lost = work done against friction

31
Q

resultant acceleration in a circle =

A

√(radial acceleration²+tangential acceleration²)

32
Q

work energy pricipal

A

work done by forces = change in KE/GPE

33
Q

coefficient of restitution, e =

A

speed of separation / speed f approach

34
Q

if e = 0

A

the two velocities after impact are the same (the particles stick together), inelastic

35
Q

If e = 1

A

the loss in KE is zero, elastic

36
Q

Component of velocity perpendicular to the line of centres

A

unaffected by collision, momentum is conserved

37
Q

Component of velocity parallel to the line of centres

A

Newtons law of restitution, momentum is conserved

38
Q

When an object hits a plane there is no impulse…

A

parallel to the plane

39
Q

Component of velocity parallel to plane

A

unaffected by collision, momentum is conserved

40
Q

Component of velocity perpendicular to plane

A

Newtons law of restitution, momentum is conserved

41
Q

newtons law of restitution

A

(Ua - Ub ) e = Vb - Va

42
Q

Impulse =

integral

A

∫Fdt

43
Q

Work done =

integral

A

∫Fdx

44
Q

Using the impulse/ work done integrals

A

find the integral and set in equal to mv - mu or Pt

45
Q

component of velocity that changes wall vs two spheres

A
  • wall: perpendicular component changes

- spheres: component parallel to line of centres changes

46
Q

stiffness =

A

λ/l

47
Q

x̄ =

A

(x1m1 + x2m2 + … + xnmn)/ (m1 + m2 + … + mn)

48
Q

ȳ =

A

(y1m1 + y2m2 + … + ynmn)/ (m1 + m2 + … + mn)

49
Q

finding CoM of compsite shaped laminas

A

in a table put each shapes area, CoM x co-ordinate, CoM y co-ordinate.
Consider x or y co-ordinates i.e. x*A … =x̄A

50
Q

If a hanging object is in equlibrium then…

A

the forces weight and tension act along the same line and there is no resultant moment.
The CoM lies directly below the point of suspension.

51
Q

volume of revolution about x-axis

A

V = ∫πy²dx

52
Q

volume of revolution about y-axis

A

V = ∫πx²dy

53
Q

Vx̄ =

A

∫πy²xdx

54
Q

Vȳ =

A

∫πx²ydy

55
Q

Finding the CoM of a solid of revolution about the x-axis:

A 
V = ∫πy²dx
Vx̄ = ∫πy²xdx
56
Q

Finding the CoM of a solid of revolution about the y-axis:

A 
V = ∫πx²dy
Vȳ = ∫πx²ydy
57
Q

Finding the angle between the point of suspension and the vertical axes

A
  • CoM directly below point of suspension so find x̄ and ȳ

- tanθ = ȳ/x̄

58
Q

CoM of a lamina using strips parallel to y-axis

A 
Ax̄ = ∫xydx
Aȳ = ∫½y²dx
59
Q

CoM of a lamina using strips parallel to the x-axis

A 
Ax̄ = ∫½x²dy
Aȳ = ∫xydy
60
Q

if an object is about to slide

A

F = μR

friction is limiting

61
Q

when the particle is at rest does friction act

A

no, there is no tendancy to slide

62
Q

an object will slide if

A

the line of action sits within the base

63
Q

an object will topple if

A

the line of action lies outside of the base

64
Q

volume of a circle

A

4/3πr^3

65
Q

volume of cone

A

h/3πr^2

66
Q

a in circular motion suvat

A

angular acceleration

AQA Mathematics (68 decks)

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